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7nadin3 [17]
3 years ago
5

What happens to the atoms that make up hydrogen fuel as it burns?

Physics
1 answer:
ale4655 [162]3 years ago
4 0

Answer:

Get turned Into Water.

Explanation:

Combustion of Hydrogen involves combining oxygen and hydrogen essentially so when oxygen and hydrogen combine water is produced, following chemical equation describes this process.

H_{2} +O_{2}=2(H_{2}O).

Resulting product is two molecules of water.

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A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radiu
anyanavicka [17]

Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

Q₁ = ∫ ρ dV

Here dV is the volume element of sphere of radius r.

Q₁ = ρ x 4π x ∫ r² dr

The limit of integration is from 0 to r as r is less than R.

Q₁ = (4π x ρ x r³ )/3

But volume charge density, ρ = \frac{3Q}{4\pi R^{3} }

So, Q_{1} = \frac{Qr^{3} }{R^{3} }

Applying Gauss law of electrostatics ;

∫ E ds = Q₁/ε₀

Here E is electric field inside the sphere and ds is surface element of sphere of radius r.

Substitute the value of Q₁ in the above equation. Hence,

E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

7 0
3 years ago
A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−
Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m.

Explanation:

Given that,

The actual wavelength of the hydrogen atom, \lambda_a=5.13\times 10^{-7}\ m

A hydrogen atom in a galaxy moving with a speed of, v=6.65\times 10^6\ m/s

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}

\lambda_o is the observed wavelength

\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m

So, the observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m. Hence, this is the required solution.

8 0
3 years ago
The amount of kinetic energy an object has depends upon which of the following factors? I. the object's velocity II. the object'
harina [27]

1.) The object's Velocity

Faster it goes, more kinetic energy it has

8 0
3 years ago
A charged object traveling 7 m in a uniform electric field of 5 N/C experiences a 4 J increase in Kinetic Energy.
Travka [436]

To solve this problem it is necessary to apply the principles of conservation of Energy in order to obtain the final work done.

The electric field in terms of the Force can be expressed as

E = \frac{F}{q} \rightarrow F=Eq

Where,

F = Force

E= Electric Field

q = Charge

Puesto que el trabajo realizado es equivalente al cambio en la energía cinetica entonces tenemos que

KE = W

KE = F*d

In the First Case,

4 = (qE)d\\q = \frac{4}{Ed}\\q = \frac{4}{5*7}\\q = 0.1142C

In Second Case,

KE = q E'd

KE = (0.1142)(40)(7)

KE = 31.976J

The total energy change would be subject to,

\Delta KE = 31.976-4

\Delta KE = 27.976J

Therefore the Kinetic Energy change of the charged object is 27.976J

3 0
3 years ago
(I will give you 10 pts) What elements are formed after a star runs out of hydrogen?
babunello [35]
<span>The core finally cools into a white dwarf, then a black dwarf. This is what happens when a normal-sized star dies. If a really huge star dies, it has so much mass that after the helium is used up, it still has enough carbon to fuse it into heavy elements like iron. When the core turns to iron, it no longer burns.

please give me </span>Brainliest answer?


4 0
3 years ago
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